= Solution
{c}
Put $C=\overline{\{z\}}$, where the <closure> is taken in the underlying <topological space> of $X$. It is nonempty because it contains $z$.
Suppose $C=A\cup B$ for two <closed subsets> of $C$ in the <subspace topology>. Since $C$ is a <closed subset> of $X$, both $A$ and $B$ are also <closed subsets> of $X$. The point $z$ lies in one of them, say $A$. By the defining minimality of the <closure>, every <closed subset> of $X$ containing $z$ contains $C$. Hence $C\subseteq A$, so $A=C$.
Thus $C$ cannot be expressed as the union of two proper <closed subsets>, proving that \b[$C$ is an irreducible closed subset]. This proves <closure of a point is irreducible> in every <topological space>; the <scheme> assumption is unnecessary. Equivalently, $z$ is a <generic point> of its own <closure>.
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